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Chapter 14: Problem 74

Indicate whether the given integral calls for integration by parts or substitution. $$\int \frac{x^{2}-3 x+1}{e^{2 x-3}} d x$$

Short Answer

Expert verified
The given integral \(\int \frac{x^{2}-3 x+1}{e^{2 x-3}} d x\) calls for integration by parts as it is a more appropriate method to solve this problem efficiently compared to substitution, helping to simplify the integrand.

Step by step solution

01

Consider substitution method

By observing the integrand, we can consider the substitution: \(u = 2x - 3\). Then, the derivative of u with respect to x is \(\frac{d u}{d x}=2\), which gives us \(d x = \frac{1}{2} d u\). We should now substitute these expressions and see if the integral simplifies: Remember that the integral is \(I=\int \frac{x^{2}-3 x+1}{e^{2 x-3}} d x\), when we use \(x=\frac{u+3}{2}\) we have \(I=\int\frac{\left(\frac{u+3}{2}\right)^2-3\left(\frac{u+3}{2}\right)+1}{e^u} \cdot \frac{1}{2} du\). Clearly, substitution does not simplify the integrand, making it more complicated instead of simplifying the integral. This makes substitution not an ideal choice. Now, let's consider applying integration by parts:
02

Consider integration by parts

To apply integration by parts, we can let \(u = x^2 - 3x + 1\) and \(dv = \frac{dx}{e^{2x-3}}\). Then, we compute the derivatives and integrals needed: \(du = (2x - 3) dx\), \(v = \int \frac{dx}{e^{2x-3}} = \frac{1}{2}\int e^{-u} du = -\frac{1}{2} e^{-u}\). Now, since we have expressed our integral as a product of u and dv, we can use integration by parts formula: \(I = uv - \int v du\). The result will still involve an integration with respect to u, but it will likely be simpler than the original integrand, making progress towards solving the integral. This makes integration by parts a more suitable method to solve this integral compared to the substitution method. So, we conclude that:
03

Conclusion

The given integral \(\int \frac{x^{2}-3 x+1}{e^{2 x-3}} d x\) calls for integration by parts as it is a more appropriate method to solve this problem efficiently compared to substitution, helping to simplify the integrand.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by Parts is a powerful technique in calculus used when dealing with the integral of a product of two functions. It's particularly useful when a direct integration is complex. This method is articulated through the formula:
\[\int u \ dv = uv - \int v \ du\]Here, the process involves choosing one part of the integrand as \(u\), which will be differentiated to \(du\), and the other part as \(dv\), which will be integrated to \(v\). When applying Integration by Parts, the goal is to simplify the integral into a more manageable form.
  • Choose \(u\) wisely: Typically, functions that become simpler upon differentiation (like logarithmic or polynomial functions) are chosen as \(u\).
  • Determine \(dv\): The remaining part of the integral is treated as \(dv\), which should be easy to integrate.
Integration by Parts is ideal when substitution complicates the integral further. In our exercise, despite trying substitution, the integration became cumbersome, making Integration by Parts the better choice.
Substitution Method
The Substitution Method is a core technique in calculus integration, serving to simplify integrals by changing variables. This technique is especially useful when you notice that a function within the integral resembles or is part of the derivative of another function. In essence, substitution can be seen as the reverse of the chain rule used in differentiation.
When using substitution:
  • Identify the substitution: Choose a part of the integrand as \(u\), such that its derivative \(du\) is also present in the integral.
  • Change variables: Substitute \(u\) and its corresponding \(du\) into the integral, often simplifying the computation.
In the examined exercise, trying the substitution method by letting \(u = 2x - 3\) did not simplify the integral, as it complicated the process further. Therefore, substitution was not the appropriate method for this integral.
Definite Integrals
Definite Integrals, unlike indefinite integrals, compute the net area under a curve between two points on the x-axis, giving a specific numerical result. They are denoted as:
\[\int_{a}^{b} f(x) \, dx\]Where "a" and "b" are the lower and upper limits, respectively. Calculating a definite integral involves evaluating the antiderivative at these limits and taking their difference:
  • Find the antiderivative \(F(x)\) of \(f(x)\).
  • Evaluate \(F(b) - F(a)\).
While our exercise focused on indefinite integrals, knowing about definite integrals is crucial for problems where bounds are set, ensuring a precise computations of areas or accumulated quantities. Understanding this can aid in comprehending the full extent of integral applications.

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Most popular questions from this chapter

Show by example that a second-order differential equation, one involving the second derivative \(y^{\prime \prime}\), usually has two arbitrary constants in its general solution.

Electric Circuits The flow of current \(i(t)\) in an electric circuit without capacitance satisfies the linear differential equation $$ L \frac{d i}{d t}+R i=V(t) $$ where \(L\) and \(R\) are constants (the inductance and resistance respectively) and \(V(t)\) is the applied voltage. (See figure.) If the voltage is supplied by a 10 -volt battery and the switch is turned on at time \(t=1\), then the voltage \(V\) is a step function that jumps from 0 to 10 at \(t=1: V(t)=5\left[1+\frac{|t-1|}{t-1}\right]\). Find the current as a function of time for \(L=R=1 .\) Use a grapher to plot the resulting current as a function of time. (Assume there is no current flowing at time \(t=0 .\) [Use the following integral formula: \(\int\left[1+\frac{|t-1|}{t-1}\right] e^{t} d t=\) \(\left.\left[1+\frac{|t-1|}{t-1}\right]\left(e^{t}-e\right)+C .\right]\)

Find the general solution of each differential equation in Exercises \(1-10 .\) Where possible, solve for \(y\) as a function of \(x\). $$\frac{d y}{d x}=\frac{1}{(x+1) y^{2}}$$

Cooling A bowl of clam chowder at \(190^{\circ} \mathrm{F}\) is placed in a room whose air temperature is \(75^{\circ} \mathrm{F}\). After 10 minutes, the soup has cooled to \(150^{\circ} \mathrm{F}\). Find the value of \(k\) in Newton's Law of Cooling, and hence find the temperature of the chowder as a function of time.

Decide whether or not the given integral converges. If the integral converges, compute its value. $$\int_{1}^{+\infty} \frac{1}{x^{1.5}} d x$$
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